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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: madscientist on February 20, 2006, 09:08:37 AM
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Hi everyone,
Lets say we are looking at an Irreversible isothermal process:
Quote, "An isothermal process is a thermodynamic process in which the temperature of the system stays constant; ?T = 0. This typically occurs when a system is in contact with an outside thermal reservoir, and the system changes slowly enough to allow it to adjust to the temperature of the reservoir. The opposite extreme in which a system exchanges no heat with its surroundings is known as an adiabatic process." end quote.
en.wikipedia.org/wiki/Isothermal
so, if an isothermal process means that Delta T =0, does that mean delta P=0 also, and if so how can the volume change.
hear is the question im working on and why im so confused by this.
A sample of 1.0 mol He (assume ideal) with a Cv = 3/2 R, initially at 298K and 10L is expanded, with the surroundings maintained at 298K, to a final volume of 20L, in two ways:
n=1mol, Cv= 12.471 JK, Tinitial = 298K, Vinitial = 10L, Vfinal = 20L
Tsurr = constant = 298K (therefore the Tsystem should be constant yes ??? )
(a.) Isothermally and reversibly
(b.) Isothermally against a constant external pressure of 0.50atm
calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.
ok, so i found it fairly easy to answer part a, and am confident with my answer:
(a.)isothermally and reversibly
wrev=-nRT ln(Vfinal/Vinitial) = -1717.322 J
* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:
q = -w = 1717.332 J
Delta S = qrev / T = 5.763 J/K
If: PiVi = PfVf = nRT
then: Delta(pV) = 0
Delta H = Delta U + Delta(pV) = 0 + 0 = 0
Delta T = 0 (isothermal)
Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q ???)
Delta A = Wrev = -1717.322 J
my working for (b.) is as follows, I get stuck at Delta T.
(b.)Isothermally against Pext = 0.50atm
(Isothermally and Irreversibly then isnt it ???)
wirrev = -Pext . Delta V = -506.6 J
Delta U = q + w = 0, (q = -w)
Delta H = Delta U + Delta (pV) = 0 + 0 = 0
Delta S = nR ln (Vfinal / Vinitial) = 5.763 J/K
now for Delta T, it should equal to zero being an isothermal process shouldnt it?
Or is it:
wirrev = -Pext . Delta V = Cv . Delta T
so: Delta T = wirrev / Cv = -40.623K
and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?
If anyone can check my work for part (a.) and the work attempted for part (b.) it would be very helpfull.
cheers,
madscientist :albert:
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so, if an isothermal process means that Delta T =0, does that mean delta P=0 also, and if so how can the volume change.
That is only true for an isoentropic (reversible & adiabatic) process
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(a.) Isothermally and reversibly
calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.
wrev=-nRT ln(Vfinal/Vinitial) = -1717.322 J
* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:
q = -w = 1717.332 J
Delta S = qrev / T = 5.763 J/K
If: PiVi = PfVf = nRT
then: Delta(pV) = 0
Delta H = Delta U + Delta(pV) = 0 + 0 = 0
Delta T = 0 (isothermal)
Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q ???)
Delta A = Wrev = -1717.322 J
Yes, you are correct for part (a)
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Cheers geodome,
I actually stuck my nose in book and did some reading,(funny how that helps ;), I didnt realise that anything with a capital, for eg dU, dT, dH.. is a state function and as long as the final conditions are the same in a reversible process as they are in an irreversible process their values are the same, their said to be path independant, whereas lower case expressions like work and heat w and q, are path dependant and will be different values for each process.
Is this reasoning correct?
cheers,
madscientist :albert:
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state functions are function whose values are path-independent.
the notation for capital is not for highlighting state functions. small capital serves to highlight the value of the state function per unit mass
eg. we use h or H to represent enthalpy. however, the unit for h is kJ/kg whereas the unit for H is kJ.
so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.
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Thats extremely handy to know geodome, Thanks heaps for all your help.
cheers,
madscientist
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Or is it:
wirrev = -Pext . Delta V = Cv . Delta T
so: Delta T = wirrev / Cv = -40.623K
and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?
We are considering an isothermal process, so this approach is wrong.
(b.) Isothermally against a constant external pressure of 0.50atm
calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.
(b.)Isothermally against Pext = 0.50atm
(Isothermally and Irreversibly then isnt it ???)
wirrev = -Pext . Delta V = -506.6 J
Delta U = q + w = 0, (q = -w)
Delta H = Delta U + Delta (pV) = 0 + 0 = 0
Delta S = nR ln (Vfinal / Vinitial) = 5.763 J/K
now for Delta T, it should equal to zero being an isothermal process shouldnt it?
We are considering the case for an isothermal & irreversible process.
dU = Q + W
since dU = n.Cv.dT where dT = 0, then dU = 0 => Q = -W
W = -pext.dV = -50650 J
Q = -W = 50650J
deltaS = n.R.ln(Vfinal/Vini) = (1.0)(8.314)ln(2) = 5.763J/mol
A = U - TS => dA = dU -TdS => deltaA = 0 - T.deltaS = -298*5.763 = 1717J
G = H - TS => dG = dH - T.dS
dH = n.Cp.dT => dH = 0 for isothermal process
dG = -T.dS => deltaG = -T.deltaS = deltaA = 1717J
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so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.
what does a unit mol consist of?, and could you repeat what you did in the last post and include units so i can understand your method a bit better.
cheers,
madscientist :albert:
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so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.
The small capital denotes the value of the state function per unit mass. mass can be expressed in either in kg or moles. w can refer to amount of work done by the system per mole of substance or per kg of substance in the system.
With regards to your question on isothermal process, we will be working on a molar basis and not kg because you are provided the molar quantity (n=1.0mol) of the gas in interest.
It actually matters if the state function is express in terms of kg or moles, because the resultant equation will not be exactly the same.
let's consider molar basis for enthalpy of a perfect gas
h = u + pv = u + RT
unit of h = J/mol
unit of u = J/mol
unit of R = J/mol.K
unit of T = K
let's consider mass basis for enthalpy of a perfect gas
h = u + pv = u + RT/M where M is the molar mass of the gas
unit of h = J/kg
unit of u = J/kg
unit of R = J/mol.K
unit of T = K
unit of M = kg/mol
unit of RT/M = J/kg
from the above example, we can see that the equations for mass and molar basis are not almost entirely the same.
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Thanks geodome
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Hi everyone,
Would like to seek your help to clarify some doubts with regards to the questions posted by madscientist.
(a) isothermal, reversible expansion of an ideal gas
Question 1: For a reversible process, shouldn't delta G = 0?
(b) isothermal expansion of an ideal gas against a constant external pressure
Question 2: In this case, delta H = 0 but q is not equal to 0.
Since the external pressure is kept constant, shouldn't delta H = qp = q?
Question 3: G is a state function. Hence the value of delta G is path-independent. For (a) and (b), are the delta G values calculated the same?
Thank you.
Regards,
eutectic6002
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(a) isothermal, reversible expansion of an ideal gas
Question 1: For a reversible process, shouldn't delta G = 0?
I think you may be confusing processes at equilibrium with reversible processes. For a reaction at equilibrium, delta G = 0. For a reversible process, delta G is not necessarily zero.
(b) isothermal expansion of an ideal gas against a constant external pressure
Question 2: In this case, delta H = 0 but q is not equal to 0.
Since the external pressure is kept constant, shouldn't delta H = qp = q?
I believe this applies only to reversible processes, since it doesn't make sense to relate a state function to a variable which is path dependent.
Question 3: G is a state function. Hence the value of delta G is path-independent. For (a) and (b), are the delta G values calculated the same?
They should be for the reason you gave.
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After some though, I have a better answer for (b):
You have to make a distinction between Pext and Pint, that is the pressure of the surroundings and the pressure of the system. delta H = qp only when the pressure of the system (Pint) stays constant. For a reversible process, Pext = Pint, so you need not make the distinction here. But, for irreversible proceses the two are not necessarily the same. In the irreversible, isothermal expansion at constant external pressure, Pext stays constant, but Pint decreases throughout the expansion, until Pext = Pint.
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Hi Yggdrasil,
Thank you for the prompt reply. Really appreciate it. I’m still quite confused and welcome more inputs from you and others.
1. Reversible processes, equilibrium and delta G
I tend to agree that delta G of reversible processes need not necessarily be zero after looking at many worked examples.
However I face difficulty explaining to students because in many textbooks and websites on Chemical Thermodynamics, it is often stated that a reversible process occurs in a series of infinitesimal steps, and at each step the system is at equilibrium with its surroundings. It is commonly stated that the entire reversible process is a succession of equilibrium states. In fact one reference even treats reversible processes as equilibrium processes (see below).
Reversible processes are also called equilibrium processes. The idea is that in a reversible process, such as a gas expansion, the system is always in balance (the external and internal pressures are only infinitesimally different) and so essentially in equilibrium. [Source: Page 21 of Energetics and Equilibria, Lecture notes from University of Cambridge]
Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.
Question 2
http://www.chem1.com/acad/webtext/thermeq/TE4.html
Problem example 2 in the above website deals with the reversible isothermal expansion of an ideal gas which is relevant to our discussion. In the given answers, delta G is not zero but delta S(universe) is zero. If delta S(universe) is zero, shouldn’t delta G = zero since delta G = –T delta S(universe)?
Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion
Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct?
2. delta H = qp
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = qp, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case, q is not qp?
Apologies for asking so much in such a messy way.
Thank you and regards,
eutectic6002
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Reversible processes are also called equilibrium processes. The idea is that in a reversible process, such as a gas expansion, the system is always in balance (the external and internal pressures are only infinitesimally different) and so essentially in equilibrium. [Source: Page 21 of Energetics and Equilibria, Lecture notes from University of Cambridge]
Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.
http://en.wikipedia.org/wiki/Quasistatic_equilibrium
Reversible processes are not necessary at equilibrium, but equilibrium is a good approximation to describe reversible processes. If these processes are really at equilibrium, then dG will actually be zero. We apply the same equations on reversible and equilibrium processes, but dG isn't always zero. Does this not indicate that reversible processes are not necessary at equilibrium?
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Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion
Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct?
What do you mean by non-expansion work? According to the Carnot Principle, the maximum work extracted from a process corresponds to the work done if the process is reversible. This maximum work is called the availability (D) of the system. The availability of system corresponds to the change of the Gibbs' free energy of the system. Since the availlability of the system is a state function, your statement is correct as long as the 3 paths share the same initial and final conditions of the gas.
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2. delta H = qp
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = qp, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case, q is not qp?
Consider the case for isothermal expansion of 1 mole of a monoatomic ideal gas:
H = U + PV
PV = RT & U = (3/2)RT
H = (5/2)RT
dH = (5/2)R.dT
Does not the above expression show that enthalpy of an ideal gas is dependent on temperature, but independent of pressure?
Since dT = 0 for isothermal processes, then dH = 0
dH = Q + W = 0
Q = -W = P.dV
This above expression means that the heat input is converted to work done corresponding to the gas expansion.
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression
How can a fixed mass of gas expand if its pressure and temperature are kept constant? (V = nRT/P)
You must keep in mind that Pexternal can be used as a substitute for Pgas if the pressure difference is negligible. It does not actually mean the gas is at constant pressure, but the value you get from using PexternaldV is a good approximation to the actual work done.
There are alot of approximations and simplification techniques (or tricks or assumptions) used in thermodynamics.
There is no Qp or Q.
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Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.
If you really think about it, it would be impossible for the gas to expand if it were always in equilibrium with its surroundings (since equilibrium implies that the state of the gas remains constant). In order to accomplish transformation reversibly, you infintesimally distrurb the equilibrium, let the system move back to equilibrium, distrub infintesimally again, etc. it is in these infintesimally pertubations of the system where you find your changes in free energy.
Question 2
http://www.chem1.com/acad/webtext/thermeq/TE4.html
Problem example 2 in the above website deals with the reversible isothermal expansion of an ideal gas which is relevant to our discussion. In the given answers, delta G is not zero but delta S(universe) is zero. If delta S(universe) is zero, shouldn’t delta G = zero since delta G = –T delta S(universe)?
delta Gsys = delta Hsys - T[sys] delta S[sys]
Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion
Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct?
2. delta H = qp
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = qp, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case, q is not qp?
Again, when delta H is defined as qp, it is assumed that the process is a reversible process, and that Pext = Pint.
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Thanks for sharing, geodome and Yggdrasil.
Still trying hard to grasp the basics of Chemical Thermodynamics.
Cheers,
eutectic6002
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Hi everyone,
Back again for clarification with regards to the answer given by Yggdrasil for Question 2 in the earlier posting.
Again please consider the reversible isothermal expansion of an ideal gas.
Good to refer to Problem example 2 in the website listed below: http://www.chem1.com/acad/webtext/thermeq/TE4.html
In this case,
delta U = 0
delta H = 0
delta Gsys < 0 and delta Ssys > 0
delta Suniverse = 0
Both the following equations are commonly quoted in textbooks and used for calculations:
Equation (1): delta Gsys = delta Hsys - T delta Ssys
Equation (2): delta Gsys = - T delta Suniverse
Equations (1) and (2) do not give the same delta G value for the above expansion process.
Shouldn't both equations give the same delta G value?
Can it be that the isothermal expansion of the ideal gas is a spontaneous process (though carried out reversibly) and hence delta Suniverse is not zero?
Another question: In this case, is pressure P considered constant?
Equation (1) is only applicable at constant P and constant T right?
Thank you once again ...
Cheers,
eutectic6002
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Here's the problem & solution from http://www.chem1.com/acad/webtext/thermeq/TE4.html for easy reference:
Problem Example 2
One mole of an ideal gas at 300 K is allowed to expand slowly and isothermally to twice its initial volume. Calculate q, w, ?S°sys, ?S°surr, and ?G° for this process.
Solution: Assume that the process occurs slowly enough that the it can be considered to take place reversibly.
a) The work done in a reversible expansion is (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.chem1.com%2Facad%2Fwebtext%2Fthermeq%2FTE-images%2Feqpe2.gif&hash=138b64347ad03360880f72e73991389a9de7297b)
so w = (1 mol)(8.314 J mol–1 K–1)(300K)(ln 2) = 1730 J.
b) In order to maintain a constant temperature, an equivalent quantity of heat must be absorbed by the system: q = 1730 J.
c) The entropy change of the system is given by Eq. 10 (Part 2): (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.chem1.com%2Facad%2Fwebtext%2Fthermeq%2FTE-images%2Feq10.gif&hash=c1e377046d21f0dbfdd14827d17456d45a2b7752)
?S°sys = R ln (V2/V1) = (8.314 J mol–1 K–1)(ln 2) = 5.76 J mol–1 K–1
d) As was explained in Part 2, the entropy change of the surroundings, formally defined as qrev/T, is
?S°surr = q/T = (–1730 J) / (300 K) = – 5.76 J mol–1 K–1.
e) The free energy change is
?G° = ?H° – T ?S°sys = 0 – (300 K)(5.76 J K–1) = –1730 J
Comment: Recall that for the expansion of an ideal gas, ?H° = 0. Note also that because the expansion is assumed to occur reversibly, the entropy changes in (c) and (d) cancel out, so that ?S°world = 0.
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Equation (2): delta Gsys = - T delta Suniverse
Shouldn't it be dHsys = - T.dSuniverse?